Show that $f(z)=\bar{z}$ is nowhere analytic

Question

Let us consider the function:

$f(z)=\bar{z}$. How can I show that this function is nowhere analytic?

It is evident that

$$\lim\limits _{z\to 0} \frac{f(z)-f(0)}{z-0}=\lim\limits_{z\to 0}\frac{\bar{z}}{z}$$

does not exist, but I am stuck to show it is not differentiable on the entire plane.

Answer 1

Perhaps easiest is to use the Cauchy-Riemann equations. $\bar{z} = x-iy$, so $u = x$ and $v=-y$ and so $u_x = 1$, $u_y = 0$, $v_x=0$, $v_y = -1$, so the equations are never satisfied, at any point.

Your technique will also work at any point. Think of the complex derivative in the alternative way: $$ \lim_{h\to 0} \frac{f(z+h)-f(z)}{h} $$